Geology Reference
In-Depth Information
To calculate the driving force, the buoyancy, note that the volume of the sinking
sheet is
V
Dd
. Because the sketch is a cross-section, the sheet
implicitly extends in the third dimension. Therefore, let us calculate the forces
on a unit length in the third dimension; that is where the factor 1 came from in
the volume. The buoyancy will depend on the average reduction in temperature
compared with the mantle. The temperature in the descending sheet is inherited
from the plate at the surface. The temperature in the surface plate is zero at the
Earth's surface and
T
at its base. Therefore, let us use the simple approximation
that the average temperature in the plate is (
T
=
D
×
d
×
1
=
+
0)
/
2
=
T/
2. Then the reduction in
temperature
T
=
(
T/
2
−
T
)
=−
T/
2. So, from Eq. (5.5), the buoyancy is
B
=−
gρ
0
αT Dd/
2
.
(5.7)
Therefore the buoyancy is negative - in other words, it is a downward force.
You may be worrying that the sheet will warm up as it descends, so we shouldn't
use a constant temperature of
T/
2 throughout its depth. However, there is a good
reason to do this. It is because the sheet will warm up by absorbing heat from the
surrounding hot mantle, so the adjacent mantle will be cooled and itself become
negatively buoyant. The effect is that the amount of heat is not changed, it is just
redistributed horizontally, so we do not make a large error by treating it as though
it is still in the descending sheet. Actually I could more accurately have said the
deficit
in heat is not changed.
Now let's consider the resisting force. This is due to the viscosity of the mantle,
so it is proportional to the relevant velocity gradient, according to the discussion
in Chapter 4. In Figure 5.2(b), the fluid on the right-hand side is descending at
velocity
v
, whereas on the left-hand side it is rising at velocity
v
. Therefore the
vertical velocity changes by the amount 2
v
over a horizontal distance
D
(assuming
the box is square, for simplicity). Thus there is a velocity gradient of 2
v/D
.This
velocity gradient is an expression of the fact that the fluid is being sheared, i.e.
deformed, in the sense of the vertical arrows in Figure 5.2(b). This deformation
will induce a stress
τ
in the fluid, which, from Eqs (4.1) and (4.3) is
τ
=
2
μv/D.
This stress acts on the descending sheet and thus generates a resisting force. The
force is the stress times the area over which it acts. For a unit length of the sheet in
the third dimension, that area is
D
×
=
1
D
. Thus the resisting force acting on the
surface of the descending sheet is
F
R
=
D
2
μv/D
=
2
μv.
This force acts upwards and is therefore positive. Actually Figure 5.2(a) suggests
that we might count resistance in other parts of the system as well. For example,
